3.58 \(\int (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=78 \[ \frac{2 b^3 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^4 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \]

[Out]

(2*b^3*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^4*Sin[c + d
*x])/(3*d*(b*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.0974828, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {16, 3012, 2642, 2641} \[ \frac{2 b^3 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^4 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(2*b^3*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^4*Sin[c + d
*x])/(3*d*(b*Cos[c + d*x])^(3/2))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=b^5 \int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac{2 A b^4 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{1}{3} \left (b^3 (A+3 C)\right ) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx\\ &=\frac{2 A b^4 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{\left (b^3 (A+3 C) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 \sqrt{b \cos (c+d x)}}\\ &=\frac{2 b^3 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^4 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.170999, size = 58, normalized size = 0.74 \[ \frac{2 b^3 \left ((A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+A \tan (c+d x)\right )}{3 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(2*b^3*((A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Tan[c + d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])

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Maple [B]  time = 3.911, size = 294, normalized size = 3.8 \begin{align*} -{\frac{2\,{b}^{3}}{3\,d} \left ( -2\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-2\,{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( A+3\,C \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+A\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +3\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) }}} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

-2/3*(-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A+3*C)*sin(1/2*d*x+1/2*c)^2+A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*b^3*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c
)^2)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2
*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + A b^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + A*b^2*cos(d*x + c)^2)*sqrt(b*cos(d*x + c))*sec(d*x + c)^5, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^5, x)